Good job, you had the first part correct and you'll use that to answer the succeeding questions.

You should know that Revenue, R = price * quantity demand.

In your equation price = p(x) and quantity demand = x.

So *R = p(x) * x*

* R...*

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Good job, you had the first part correct and you'll use that to answer the succeeding questions.

You should know that Revenue, R = price * quantity demand.

In your equation price = p(x) and quantity demand = x.

So *R = p(x) * x*

*R = (-0.1x + 525) * x*

* R = -0.1x^2 + 525x*

to maximize the revenue, get the first derivative in terms of x

R' = -0.1*2 *x + 525

set R' = 0

0 = -0.2 x + 525

solve for x

**x = 2625 **--------> this is the quantity demand that will give you maximum revenue. you can check it if it is really the value that'll give the maximum revenue by: p(x) = -0.1(2625) + 525

R = p(x) (2625)

after that, get a value lower or higher than 2625, say 2624 and 2626, you'll see that the R will not be higher that R of 2625.

so then, use **x=2625 in p(x) = -0.1x + 525**

** p(x) = 262.5 **-----> this value is the original value minus the rebate.

To answer the second question, you just subtract 262.5 from 420.

so rebate = 420 - 262.5 = 157.5

and lastly (it seems that you did not write the function), I assumed that it is the same as the first part. So to have the maximum profit, the revenue must be maximum. Since we already computed the rebate for maximum revenue which is 157.5, the size is157.5/420 * 100 =

**37.5% **of the original value....

Hope this helps :)